LeetCode: 109. Convert Sorted List to Binary Search Tree

题目描述

Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

Example:

Given the sorted linked list: [-10,-3,0,5,9],
One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:

      0
 / \
   -3 9  / /
 -10 5

题目大意: 根据给定的有序链表，生成一种可能的平衡二叉树。

解题思路

基本思路同 LeetCode: 108. Convert Sorted Array to Binary Search Tree 题解。由于给定数据结构是链表，因此只能先生成左子树，以使迭代指针指向链表中根节点的元素。

AC 代码

/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */
class Solution {
private:
    // 将 head 前 n 个元素生成平衡二叉树 BST
    TreeNode* sortedListToBSTRef(ListNode*& head, int n)
    {
        if(head == nullptr || n <= 0) return nullptr;

        // 生成左子树
        TreeNode* leftChildTree = sortedListToBSTRef(head, n/2);

        // 生成根节点
        TreeNode* root = new TreeNode(head->val);
        head = head->next;

        // 生成右子树
        TreeNode* rightChildTree = sortedListToBSTRef(head, n - n/2 - 1);

        // 链接左/右子树
        root->left = leftChildTree;
        root->right = rightChildTree;

        return root;
    }
public:
    TreeNode* sortedListToBST(ListNode* head) {
        int n = 0; //记录节点个数

        // 计算节点个数
        ListNode* iter = head;
        while(iter != nullptr && ++n) iter = iter->next;

        return sortedListToBSTRef(head, n);
    }
};